Re: [xsl] Replace newline characters(\n) with <br>

["Joerg Heinicke" <joerg dot heinicke at gmx dot de>]

You can search the archives for 'replace' or 'recursive templates', such
questions are quite often.

<xsl:template match="message">
    <xsl:call-template name="replace">
        <xsl:with-param name="string" select="."/>
    </xsl:call-template>
</xsl:template>

<xsl:template name="replace">
    <xsl:param name="string"/>
    <xsl:choose>
        <xsl:when test="contains($string,'&#10;')">
            <xsl:value-of select="substring-before($string,'&#10;')"/>
            <br/>
            <xsl:call-template name="replace">
                <xsl:with-param name="string"
select="substring-after($string,'&#10;')"/>
            </xsl.call-template>
        </xsl:when>
        <xsl:otherwise>
            <xsl:value-of select="$string"/>
        </xsl:otherwise>
    </xsl:choose>
</xsl:template>

Regards,

Joerg


> Hello,
>
>   I have the a simple Xml document as shown below
>
> <xml version="1.0">
> <message>
> string1
> string2
> string3
> </Message>
>
> Technically, the strings are separated by newlines. Now when i write an
XSL
> for this document, i want to replace all the newlines with <br> so that
> these strings appear in three lines in the output.
>
> Thanks,
>
>
> Arun Ramadoss
> Professional
> Software Engineer II
> E*TRADE Financial
> w 650-331-6307


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